3/(x-5)+8/x=2 equation

Given the equation:
$$\frac{3}{x - 5} + \frac{8}{x} = 2$$
Multiply the equation sides by the denominators:
x and -5 + x
we get:
$$x \left(\frac{3}{x - 5} + \frac{8}{x}\right) = 2 x$$
$$\frac{11 x - 40}{x - 5} = 2 x$$
$$\frac{11 x - 40}{x - 5} \left(x - 5\right) = 2 x \left(x - 5\right)$$
$$11 x - 40 = 2 x^{2} - 10 x$$
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$11 x - 40 = 2 x^{2} - 10 x$$
to
$$- 2 x^{2} + 21 x - 40 = 0$$
This equation is of the form

a*x^2 + b*x + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -2$$
$$b = 21$$
$$c = -40$$
, then

D = b^2 - 4 * a * c = 

(21)^2 - 4 * (-2) * (-40) = 121

Because D > 0, then the equation has two roots.

x1 = (-b + sqrt(D)) / (2*a)

x2 = (-b - sqrt(D)) / (2*a)

or
$$x_{1} = \frac{5}{2}$$
$$x_{2} = 8$$